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The base case is obvious, as $|a_{1}| \leq |a_{1}|$ is true.

The inductive case is where I'm having trouble. Here is what I have: Suppose $|a_{1} + a_{2} + \dots + a_{n}| \leq |a_{1}| + |a_{2}| + \dots + |a_{n}|$ is true for $n\in \mathbb{N}$. From this, $|a_{1} + (a_{2} + \dots + a_{n} + a_{n+1})| \leq |a_{1}| + |a_{2} + \dots + a_{n} + a_{n+1}|$.

How can I show that if you keep applying the additive associative property of fields repeatedly to the inside part of the absolute value, then you will eventually show the original proposition?

Any help would be greatly appreciated. Thank you.

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    Now assume $\vert a_1 + a_2 \vert$ $\leq$ $\vert a_1 \vert$ + $\vert a_2 \vert$ is true. Use this to prove that this equality holds for more than 2 terms. – 2017-02-17
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    Thank you! I see how to apply the triangle inequality and the placement of the parentheses. – 2017-02-17
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    $a_2 .... a_{n+1}$ is only $n$ terms so $|a_2 + .... + a_{n-1}| \le |a_2| + .....|a_{n+1}|$. The actual values or labels of the variables don't matter. But if you put the parenthesis around the *first* terms instead of the last it's visually much more obvious. – 2017-02-17

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If your "from this" step is valid, couldn't you do this: $$\begin{align} |(a_{1} + a_{2} + \dots + a_{n}) + a_{n+1}| & \leq |a_{1} + a_{2} + \dots + a_{n}| + |a_{n+1}| \\ &\leq |a_{1}| + |a_{2}| + \dots + |a_{n}| + |a_{n+1}| \end{align}$$ with the last step being from the inductive hypothesis?

Something seems fishy about the first step though, unless you can consider it a case where n=2?

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    Good point! Then base case $n = 1$ won't give us the inductive case. We need a base case of $n = 2$. Other wise we have an all horses are the same color issue. – 2017-02-17
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Type instead $$|(a_1+a_2+\cdots + a_n)+a_{n+1}|\leq |a_1+\cdots + a_n|+|a_{n+1}|$$ And now use the inductive step to conclude

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    I see my error. I was applying the parentheses around the wrong set of terms. Thank you for your help! – 2017-02-17
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    The parenthises don't matter as $a_2.... a_{n+1}$ also only have $n$ terms and you are assuming true for $n$ terms. But putting the parenthesis around the first terms rather than the last will sure make things clearer and more intuitive for the prover to see. – 2017-02-17
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$|a_1 + a_2 + ..... + a_n + a_{n+1}| = |(a_1 + a_2 + ..... + a_n) + a_{n+1}|\le$

$|a_1 + a_2 + ..... + a_n)| + |a_{n+1}|$ (as those are only two terms)

$\le (|a_1| + |a_2| + ..... + |a_n|) + |a_{n+1}|$ (as those are $n$ terms)

$= |a_1| + |a_2| + ..... + |a_n| + |a_{n+1}|$

P.S.

"How can I show that if you keep applying the additive associative property of fields repeatedly to the inside part of the absolute value, then you will eventually show the original proposition?"

You don't have to.

$|a_1 + ..... + a_n| \le |a_1| + |a_2| + ..... + |a_n|$ because there are $n$ terms.

So

$|a_2 + ....... + a_{n+1} | \le |a_2| + |a_3|+ ..... +|a_{n+1}|$ because there are $n$ terms.

P.S.S (!!!IMPORTANT!!!)

Because the induction case uses the hypotheses on two terms, the base case that $n = 1$ won't do it.

This leads to all "all horses are the same color" paradox.

Consider $|a_1 + .... a_n| \ge |a_1| + .... + |a_n|$ (Not true).

True for $n = 1$: As $|a_1| \ge |a_1|$.

And True for induction as $|(a_1+ ..... + a_n) + a_{n+1}| \ge |a_1+ .... + a_n| + |a_{n+1}| \ge |a_1|+..... + |a_n| + |a_{n+1}|$

But it isn't true as we never showed it was true for $n = 2$.

So base case is you must show $|a + b| \le |a| + |b|$ can you do that?

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    That is an *excellent* p.s.! – 2017-02-17